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Chapter 6: Problem 26

Consider an ergodic \(M / M / s\) queue in steady state (that is, after a long time) and argue that the number presently in the system is independent of the sequence of past departure times. That is, for instance, knowing that there have been departures \(2,3,5\), and 10 time units ago does not affect the distribution of the number presently in the system.

Short Answer

Expert verified
In an ergodic and steady-state M/M/s queueing system, both arrivals and service times follow exponential distributions, which are memoryless. The system can be modeled as a continuous-time Markov chain, which means that the future state of the system depends only on the present state and not on the past states. According to the Markov chain property, the number of individuals currently in the system depends solely on the stochastic characteristics of arrivals and service times, and not on historical data such as previous departure times. Consequently, the distribution of the present number in the system is independent of the sequence of past departure times.

Step by step solution

01

Understanding ergodic and steady-state M/M/s queue

Ergodicity in queueing theory means the system has reached a stable state, where every state that the process can enter will be visited infinitely many times. This indicates that the statistics of the system, such as the distribution of customers, do not change over time. In other words, the system entered its steady-state.
02

Describe the inter-arrival and service time distribution

The inter-arrival times follow an exponential distribution with a rate parameter 位. The service times also follow an exponential distribution with a rate parameter 碌. Both the inter-arrival and service times are memoryless, meaning that they do not depend on their past values.
03

Relate the M/M/s queue to a continuous Markov chain

By using the Poisson process for arrivals and exponential distribution for service times, we can model the M/M/s queue as a continuous-time Markov chain, where the state transitions (arrivals and departures) take place according to the service rate 碌 and the arrival rate 位.
04

Explain the Markov chain property

The Markov chain property states that the future state of the system is only dependent on the present state, and not on the past states. In the context of our queueing system, it means that the number of customers in the system at a given time does not depend on the prior states, such as past departure times.
05

Argue the independence of present number in the system and past departure times

Since the M/M/s queue can be modeled as a continuous Markov chain, and its steady-state distribution is achieved, we have the Markov chain property. Therefore, the number of individuals currently in the system depends only on the stochastic characteristics of arrivals and service times rather than historical data such as previous departure times. As a result, the distribution of the present number in the system is independent of the sequence of past departures.

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Most popular questions from this chapter

Individuals join a club in accordance with a Poisson process with rate \(\lambda\). Each new member must pass through \(k\) consecutive stages to become a full member of the club. The time it takes to pass through each stage is exponentially distributed with rate \(\mu .\) Let \(N_{i}(t)\) denote the number of club members at time \(t\) who have passed through exactly \(i\) stages, \(i=1, \ldots, k-1 .\) Also, let \(\mathbf{N}(t)=\left(N_{1}(t), N_{2}(t), \ldots, N_{k-1}(t)\right)\) (a) Is \(\\{\mathbf{N}(t), t \geqslant 0\\}\) a continuous-time Markov chain? (b) If so, give the infinitesimal transition rates. That is, for any state \(\mathbf{n}=\) \(\left(n_{1}, \ldots, n_{k-1}\right)\) give the possible next states along with their infinitesimal rates.

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